# Lagrange multiplier problem of looking for the point on $\frac1x + \frac1y + \frac1z =1$ closest to the origin

Use Lagrange multipliers to find the point on the surface $$\frac1x + \frac1y + \frac1z =1$$ which is closest to the origin.

I was wondering if I would start off by using the distance formula, $$d=\sqrt{\left(\frac1x\right)^2+\left(\frac1y\right)^2+\left(\frac1z\right)^2}$$ which would then simplify to, $$d^2=\left(\frac1x\right)^2+\left(\frac1y\right)^2+\left(\frac1z\right)^2$$ I don't know where to go from there, so if someone could help me out, that would be great.

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Clearly you should re-read the section on lagrange multipliers, but I'll remind you briefly: If you have a function $f$ to minimize with constraint $g = 0$, then you should minimize $f - \lambda g$. In your case, the function you want to minimize is distance: $d^2 = x^2+y^2+z^2$. The constraint is $\frac{1}{x} +\frac{1}{y} + \frac{1}{z} -1 = 0$. – user81327 Jul 19 '13 at 0:25

This is not strictly speaking Lagrange multipliers, but the idea is the same. To me, this gives a better view of what is going on.

You want the point to be on the surface, so we only want to look at the variations so that $$\frac{\delta x}{x^2}+\frac{\delta y}{y^2}+\frac{\delta z}{z^2}=0$$ Furthermore, we want to minimize $x^2+y^2+z^2$, so we want to find the point on the surface where $$x\,\delta x+y\,\delta y+z\,\delta z=0$$ The fundamental theorem of the calculus of variations says that for the all variations orthogonal to $\left(\frac1{x^2},\frac1{y^2},\frac1{z^2}\right)$ to be orthogonal to $(x,y,z)$, there must be a constant so that $$k\left(\frac1{x^2},\frac1{y^2},\frac1{z^2}\right)=(x,y,z)$$ This implies that $x^3=y^3=z^3=k$. Therefore, the point is $(3,3,3)$.

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I don't have time to work out all the details in an answer, but here's a quick starter.

The key idea behind Lagrange multipliers is that when two surfaces are tangent to each other, their normal vectors at that point are parallel. In this case, you want to find when the surface $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1$ is closest to the origin, that is, when it is tangent to a sphere of some radius.

Note first that if $F(x,y,z) = \frac{1}{x} + \frac{1}{y} + \frac{1}{z}$, the surface is equal to $F^{-1}(1)$, so the normal field is given by the gradient $\nabla F$.

It's easier to use the energy function $d^2(x,y,z) = x^2 + y^2 + z^2$, because square roots are silly. To find where distance from the origin is minimized, find the gradient of the distance function, $\nabla (d^2)$, and find where it points the same direction as $\nabla F$: $$\nabla(d^2) = \lambda \nabla F,$$ subject to $F = 1$.

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A quick and cheap way to do this problem is to exploit the symmetry. Using Lagrange's method will yield a system of equations that is symmetric with respect to any permutation of $x,y,$ and $z$, so we can assume that $x=y=z$. It follows that they are all equal to $3$.

To actually use the method, setting $\nabla(d^2)=\lambda\nabla g$ gives:

$$2x=-\lambda\frac{1}{x^2}\\2y=-\lambda\frac{1}{y^2}\\2z=-\lambda\frac{1}{z^2}\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$$

If you solve these equations, you will recover the result above.

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